16-Civ-B3 Geotechnical Design · December 2017

Question 4 of 9: Plate load test in saturated clay — proof of size independence

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Question 4: Plate load test in saturated clay — proof of size independence (7 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Terzaghi's equation $q_{ult}=cN_c+qN_q+0.5\gamma BN_\gamma$; a saturated clay loaded rapidly, so the response is undrained; a test plate of width $B_P$ and a prototype foundation of width $B_F$, both bearing on the same clay.

Find. Show that $q_{ult}$ contains no term in $B$ under undrained conditions, so that $(q_{ult})_P=(q_{ult})_F$ whatever the ratio $B_F/B_P$, and state the assumptions on which that result rests.

Approach. Evaluate Terzaghi's three bearing-capacity factors at the undrained friction angle $\phi_u=0$, then inspect which of the three terms can depend on the footing width.

1. Set the undrained strength parameters. A saturated clay sheared fast enough that no drainage occurs deforms at constant volume, so the total-stress envelope is horizontal: $$\phi_u = 0, \qquad c = c_u$$ This is the only physical input the proof needs; everything else is arithmetic on Terzaghi's factors.
2. Evaluate the surcharge factor. Terzaghi's $N_q=\dfrac{e^{2\left(3\pi/4-\phi'/2\right)\tan\phi'}}{2\cos^{2}\!\left(45^{\circ}+\phi'/2\right)}$. Substituting $\phi'=0$ makes the exponent zero, so the numerator is unity, while $\cos^{2}45^{\circ}=0.5$, giving $$N_q=\frac{1}{2(0.5)}=1$$
3. Evaluate the self-weight factor. $N_\gamma$ measures the contribution of the weight of the soil inside the failure wedge, and it is generated entirely by frictional resistance along the log-spiral. Vesic's closed form makes this explicit: $N_\gamma=2(N_q+1)\tan\phi'$, and with $\phi'=0$ the factor $\tan\phi'$ vanishes: $$N_\gamma = 2(1+1)\tan 0^{\circ} = \boxed{0}$$ Terzaghi's own tabulated $N_\gamma$ likewise goes to zero at $\phi'=0$. This is the pivot of the whole proof, because $N_\gamma$ is the only factor that multiplies $B$.
4. Evaluate the cohesion factor. Terzaghi's $N_c=(N_q-1)\cot\phi'$ is indeterminate at $\phi'=0$; taking the limit gives Prandtl's value $N_c=\pi+2=5.14$ for a smooth strip, and Terzaghi's rough-base strip value, which is the one tabulated with his equation, is $$N_c = 5.7$$ Either way it is a pure number, independent of $B$.
5. Assemble the equation. Substituting the three factors, $$q_{ult}=c_u(5.7)+q(1)+0.5\,\gamma B(0)=5.7\,c_u+\gamma D_f$$ and if the test and the foundation are both taken at the same founding level the surcharge $q=\gamma D_f$ is identical for the two, so $$\boxed{(q_{ult})_P=(q_{ult})_F=5.7\,c_u+\gamma D_f}$$ For a plate test at the ground surface, $D_f=0$ and $q_{ult}=5.7c_u$ exactly.
6. Read the result. Neither surviving term contains $B$. The width entered Terzaghi's equation only through the self-weight term, and that term is switched off by the absence of friction, so the ultimate bearing capacity of a 300 mm test plate and of a 3 m footing on the same saturated clay are numerically equal. Contrast this with a sand, where $\phi'$ is large, $N_\gamma$ is of order 30 to 80, and the same comparison would be meaningless.